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notka56 [123]
3 years ago
5

What is the volume of the square pyramid shown below?

Mathematics
2 answers:
vfiekz [6]3 years ago
8 0
The volume of a square pyramid is given by the following formula:

V = a^{2} \frac{h}{3}

a is the side length of the square base, and h is the height of the pyramid.

Plug in your known values into the formula:

a = 9, h = 5
V = 9^{2} \times \frac{5}{3}
V = 81 \times \frac{5}{3}
V = 135

The volume of this square pyramid is 135 centimeters cubed.

natta225 [31]3 years ago
4 0
The answer is 135 cm cubed.

The equation is L * W * (H/3) = volume.
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In order to determine the height of the flagpole in the school yard, Cindy is going to use similar triangles. The
kvv77 [185]

The height of the pole is 12.5 feet.

<h3>What is Unitary method?</h3>

The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

Given:

Cindy's shadow is 5 feet.

length of the shadow of the pole at the same time, she finds it to be 12.5 feet.

So,

12.5/5=x/5,

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Hence, the height of the pole is 12.5 feet.

Learn more about this Unitary method here:

brainly.com/question/22056199

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4 0
2 years ago
Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.
Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

4 0
3 years ago
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